Question: $f(n)=45\cdot\left(\dfrac{4}{5}\right)^{{\,n-1}}$ Complete the recursive formula of $f(n)$. $f(1)=$
Answer: From the explicit formula, ${45}\cdot\left({\dfrac45}\right)^{n-1}$, we can tell that the first term of the sequence is ${45}$ and the common ratio is ${\dfrac45}$. This is the recursive formula of the sequence: $\begin{cases} f(1)={45} \\\\ f(n)=f(n-1)\cdot{\dfrac45} \end{cases}$